Heat exchanger calculation
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The operating pressure of the heat exchanger is 25 MPa, the inlet temperature is 25°C, and the outlet temperature is 550°C. The flow rate is 750 kg/h. Working fluid: water. Calculate the required heat. It’s better to include formulas; I’m a beginner and want to learn*.
Q=KMΔT; water at 550 degrees should vaporize at 25 Mpa, right?
Q=750x525x4.2
Q = (H2 – H1) × W × 3600 (kW), where H2 is the enthalpy value at 25 MPa and 550°C (kJ/kg); H1 is the enthalpy value at 25 MPa and 25°C (kJ/kg); W is the water flow rate (kg/h)
As mentioned above, I did the calculations: at 25 degrees, the enthalpy of unsaturated water is 127.76 kJ/kg, while at 550 degrees it is superheated steam with an enthalpy of 3339.28 kJ/kg. Its heat value is 2,440,755 kJ, which is equivalent to 678 kW/h.
A simple calculation using the latent heat of vaporization: Q=750/3600*((550-25)*4.18+1564)=783KW
My algorithm is the same as the one on floor 4, and it includes the latent heat of vaporization component
H2 should be the enthalpy of steam； That includes the latent heat
Support for the fourth floor: I think in cases where there is latent heat of vaporization, enthalpy is the only parameter that can be used for calculation
I think it is more appropriate to use enthalpy values for calculation